Question: Is ${876267}$ divisible by $9$ ?
A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {876267}= &&{8}\cdot100000+ \\&&{7}\cdot10000+ \\&&{6}\cdot1000+ \\&&{2}\cdot100+ \\&&{6}\cdot10+ \\&&{7}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {876267}= &&{8}(99999+1)+ \\&&{7}(9999+1)+ \\&&{6}(999+1)+ \\&&{2}(99+1)+ \\&&{6}(9+1)+ \\&&{7} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {876267}= &&\gray{8\cdot99999}+ \\&&\gray{7\cdot9999}+ \\&&\gray{6\cdot999}+ \\&&\gray{2\cdot99}+ \\&&\gray{6\cdot9}+ \\&& {8}+{7}+{6}+{2}+{6}+{7} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${876267}$ is divisible by $9$ if ${ 8}+{7}+{6}+{2}+{6}+{7}$ is divisible by $9$ Add the digits of ${876267}$ $ {8}+{7}+{6}+{2}+{6}+{7} = {36} $ If ${36}$ is divisible by $9$ , then ${876267}$ must also be divisible by $9$ ${36}$ is divisible by $9$, therefore ${876267}$ must also be divisible by $9$.